Thread: Potential difference

If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

Ex=d(2x^2)/dx=4x=12

Ey=d(-2y^2)/dy=-4y=-12

Therefore, E={12i - 12j}

Not entirely sure, this is the first of its kind.

Originally Posted by quantoembryo

If we're given V=2x^2+2y^2 and we're told to find the field at a point x=3, y=3, would this method be correct?

Ex=d(2x^2)/dx=4x=12

Ey=d(-2y^2)/dy=-4y=-12

Therefore, E={12i - 12j}

Not entirely sure, this is the first of its kind.

Mostly correct. so you are off by a minus sign.

-Dan

So then it would be E={-12i+12j}? That actually makes a lot more sense.

Do you have a typo in your potential?

You wrote in your post that:

You keep responding as if

Good catch. I missed that.

-Dan

Yes, that was a typo. Sorry! Thanks for the help! One quick question pertaining to that relationship you posted. What is that greek symbol and what does it mean? Thanks

If you are referring to the in the topsquark's expression then it's not a Greek letter, actually. It's called the "del" operator, or also the "nabla". In this context, it indicates that you are to take the gradient of the scalar function

Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?

Alright, I have yet to encounter that sign so it's probably best if I say E=-dv/dx correct?

Not true for your problem, actually. If you were working in only one dimension, it would be true. For your case, you should have the vector equation



Or, for two dimensions, you'd have

Alright, let me rephrase. Ex=-dV/dx, and Ey=-dV/dy

As long as your "d's" mean "partial derivative" and not "ordinary derivative", you're ok there.